# How do you use the definition of the derivative to find the derivative of the function f(x) = 6 x + 2sqrt{x}?

Apr 9, 2018

#### Explanation:

Defintion of a derivative:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$f \left(x\right) = 6 x + 2 \sqrt{x}$
$f \left(x + h\right) = 6 \left(x + h\right) + 2 \sqrt{x + h}$

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{6 \left(x + h\right) + 2 \sqrt{x + h} - \left(6 x + 2 \sqrt{x}\right)}{h}$

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{6 h + 2 \left(\sqrt{x + h} - \sqrt{x}\right)}{h}$

$f ' \left(x\right) = 6 + 2 {\lim}_{h \rightarrow 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$

$f ' \left(x\right) = 6 + 2 {\lim}_{h \rightarrow 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \times \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}$

$f ' \left(x\right) = 6 + 2 {\lim}_{h \rightarrow 0} \frac{x + h - x}{h \left(\sqrt{x + h} + \sqrt{x}\right)}$

$f ' \left(x\right) = 6 + 2 {\lim}_{h \rightarrow 0} \frac{h}{h \left(\sqrt{x + h} + \sqrt{x}\right)}$

$f ' \left(x\right) = 6 + 2 {\lim}_{h \rightarrow 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}$

$f ' \left(x\right) = 6 + 2 \times \frac{1}{\sqrt{x} + \sqrt{x}}$

$f ' \left(x\right) = 6 + \frac{1}{\sqrt{x}}$